AIME 2009 I · 第 11 题

Solution 1 (Matrix's, Determinants)

Let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$

We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem).

$\det \left(\begin{array}{c} P \\ Q\end{array}\right)=\det \left(\begin{array}{cc}x_1 &y_1\\x_2&y_2\end{array}\right).$

Since the triangle has half the area of the parallelogram, we just need the determinant to be even.

The determinant is

$$(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))$$ Since $2009$ is not even, $x_1-x_2$ must be even, thus the two $x$'s must be of the same parity. Also note that the maximum value for $x$ is $49$ and the minimum is $0$. There are $25$ even and $25$ odd numbers available for use as coordinates and thus there are $(_{25}C_2)+(_{25}C_2)=\boxed{600}$ such triangles.

Solution 2

As in Solution 1, let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$. Let the line $41x+y=2009$ intersect the x-axis at $X$ and the y-axis at $Y$. $X$ has coordinates $(49,0)$, and $Y$ has coordinates $(0,2009)$. As such, there are exactly $50$ lattice points on this line that can be used for $P$ and $Q$.

WLOG, let the x-coordinate of $P$ be less than the x-coordinate of $Q$. Note that $[OPQ]=[OYX]-[OYP]-[OXQ]$. We know that $OY=2009$ and $OX= 49$; as such, $[OYX]=\frac{1}{2} \cdot OY \cdot OX = \frac{1}{2} \cdot 2009 \cdot 49$. In addition, $[OYP]=\frac{1}{2} \cdot 2009 \cdot x_1$ and $[OXQ]=\frac{1}{2} \cdot 49 \cdot y_2$.

Since $2009 \cdot 49$ is odd, the total area of $OYX$ is not an integer; rather, it is of the form $k + \frac{1}{2}$ where $k$ is an integer. To ensure $[OPQ]$ has an integral value, exactly one of $[OPY]$ and $[OQX]$ must have an integral value as well (the other must be of the form $k + \frac{1}{2}$ where $k$ is an integer).

Returning to $41x+y=2009$, we notice that integer pairs of $x$ and $y$ that satisfy the equation always have different parities. To satisfy exactly one of $[OPY]$ and $[OQX]$ having an integral area, we must have $x_1$ and $y_2$ having different parities. This is because having an even number for $x_1$ or $y_2$ makes the area of the triangle an integer. We can therefore deduce that $x_1$ and $x_2$ have the same parity.

Out of the $50$ usable lattice points for $P$ and $Q$, $25$ have even x-coordinates and $25$ have odd x-coordinates. Since we must pick two points with even x-coordinates or two points with odd x-coordinates, our desired answer is $\binom{25}{2}+\binom{25}{2}=300+300=\fbox{600}$.

Solution 3 (Analytic geometry)

As in the solution above, let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$.

If the coordinates of $P$ and $Q$ have nonnegative integer coordinates, $P$ and $Q$ must be lattice points either

• on the nonnegative x-axis

• on the nonnegative y-axis

• in the first quadrant

We can calculate the y-intercept of the line $41x+y=2009$ to be $(0,2009)$ and the x-intercept to be $(49,0)$.

Using the point-to-line distance formula, we can calculate the height of $\triangle OPQ$ from vertex $O$ (the origin) to be:

$\dfrac{|41(0) + 1(0) - 2009|}{\sqrt{41^2 + 1^2}} = \dfrac{2009}{\sqrt{1682}} = \dfrac{2009}{29\sqrt2}$

Let $b$ be the base of the triangle that is part of the line $41x+y=2009$.

The area is calculated as: $\dfrac{1}{2}\times b \times \dfrac{2009}{29\sqrt2} = \dfrac{2009}{58\sqrt2}\times b$

Let the numerical area of the triangle be $k$.

So, $k = \dfrac{2009}{58\sqrt2}\times b$

We know that $k$ is an integer. So, $b = 58\sqrt2 \times z$, where $z$ is also an integer.

We defined the points $P$ and $Q$ as $P=(x_1,y_1)$ and $Q=(x_2,y_2)$.

Changing the y-coordinates to be in terms of x, we get:

$P=(x_1,2009-41x_1)$ and $Q=(x_2,2009-41x_2)$.

The distance between them equals $b$.

Using the distance formula, we get

$PQ = b = \sqrt{(-41x_2+ 41x_1)^2 + (x_2 - x_1)^2} = 29\sqrt2 \times |x_2 - x_1| = 58\sqrt2\times z$ $(*)$

WLOG, we can assume that $x_2 > x_1$.

Taking the last two equalities from the $(*)$ string of equalities and putting in our assumption that $x_2>x_1$, we get

$29\sqrt2\times (x_2-x_1) = 58\sqrt2\times z$.

Dividing both sides by $29\sqrt2$, we get

$x_2-x_1 = 2z$

As we mentioned, $z$ is an integer, so $x_2-x_1$ is an even integer. Also, $x_2$ and $x_1$ are both positive integers. So, $x_2$ and $x_1$ are between 0 and 49, inclusive. Remember, $x_2>x_1$ as well.

• There are 48 ordered pairs $(x_2,x_1)$ such that their positive difference is 2.

• There are 46 ordered pairs $(x_2,x_1)$ such that their positive difference is 4.

...

• Finally, there are 2 ordered pairs $(x_2,x_1)$ such that their positive difference is 48.

Summing them up, we get that there are $2+4+\dots + 48 = \boxed{600}$ triangles.

Solution 4

We present a non-analytic solution; consider the lattice points on the line $41x+y=2009$. The line has intercepts $(0, 2009)$ and $(49, 0)$, so the lattice points for $x=0, 1, \ldots, 49$ divide the line into $49$ equal segments. Call the area of the large triangle $A$. Any triangle formed with the origin having a base of one of these segments has area $A/49$ (call this value $B$) because the height is the same as that of large triangle, and the bases are in the ratio $1:49$. A segment comprised of $n$ small segments (all adjacent to each other) will have area $nB$. Rewriting in terms of the original area, $A=(\frac{1}{2})(49)(2009)$, $B=\frac{2009}{2}$, and $nB=n(\frac{2009}{2})$. It is clear that in order to have a nonnegative integer for $nB$ as desired, $n$ must be even. This is equivalent to finding the number of ways to choose two distinct $x$-values $x_1$ and $x_2$ ($0 \leq x_1, x_2 \leq 49$) such that their positive difference ($n$) is even. Follow one of the previous methods above to choose these pairs and arrive at the answer of 600.

Solution 5 (Shoelace Theorem)

Label point $P$ and $Q$ as $(x_1, 2009 - 41x_1)$ and $(x_2, 2009 - 41x_2)$ respectively. Now, we shall use Shoelace Theorem to calculate the area of the triangle. After doing the calculations and simplifications with the coordinates, you will get the area of the triangle to be $\frac{2009|x_1-x_2|}{2}$. Based on this, in order for the area of the triangle to be an integer, $|x_1-x_2|$ needs to be even. The only way for this to happen is that either $x_1$ and $x_2$ are both even are both odd. From the equation from the problem, we can know that $0 \le x_1, x_2 \le 49$. There are $_{25}C_2$ ways to choose 2 even integers in that range or 2 odd numbers in that range. Thus, the answer is $2(_{25}C_2)$ which is $\boxed{600}$.

~ROGER8432V3

Solution 6 (Easier Shoelace)

Dividing by 41 on both sides yields

$$x+ \frac{y}{41} = 49$$ and because the result is an integer, $\frac{y}{41} = z$ and thus we have

$$x+z = 49$$ and through stars-and-bars the number of ways to count the following integer pairs greater than or equal to 0 is simply just $\binom{50}{1} = 50$.

From here, shoelace on the coordinates $(0,0)$, $P(x_1, y_1)$, $Q(x_2,y_2)$ results in $\frac{1}{2} | x_1y_2 - y_1x_2 |$. Also notice that for the resulting area to be an integer we must have that $x_1y_2 \equiv y_1x_2 \pmod{2}$. The only way this is possible is if both $x_1y_2 - y_1x_2$ are either even or odd. There are 25 pairs of odd numbers, and 25 pairs of even numbers out of the 50 possible stars-and-bars cases. Note that choosing the two numbers themselves to give to the coordinates of $P$ and $Q$ respectively make the triangle distinct as order does not matter. Thus the answer is just the number of ways to choose two even coordinates for one of the four distinct values and the remainder is for one of the odd coordinates, and thus our answer is just $2(\binom{25}{2}) = \boxed{600}$.

~Pinotation